Transcript

Indian Institute of Technology Madras Chennai, India.

Structural Engineering Division Department of Civil Engineering

1

Base - Excited SDF Linear Systems
The response of the viscously damped linear (system) oscillator shown in the sketch will now be investigated for an excitation of the base. k c y(t) Displacement of the base at any time ‘t’ will be denoted by y(t) and the associated velocity and acceleration will be denoted by y (t ) and y (t ) respectively. The exciting motion is considered to be known and unaffected by the motion of the oscillator itself. As before, the absolute displacement of the mass will be denoted by X and associated velocity and acceleration by x(t ) and x(t ) Both X and Y are measured from the static equilibrium, and both they and their derivatives are considered to be positive when directed to the right. x(t)

m

P(t)

2

The equation of motion for the system is obtained as usual by considering the equilibrium of forces acting on the mass. These forces include the spring force, damping force and D’Alembert inertia force all of each are directed to the left. Equilibrium requires that Equilibrium requires that m x + c ( x − y ) + k ( x − y ) = 0 (B1)

The equation can now be written either in terms of the absolute displacement x, as m x + c x + kx = k y ( t ) + c y ( t ) (B2) or in terms of relative displacement, or spring deformation, u= x-y (B3) (B4) m u + c u + ku = − m y (t ) Upon dividing through by ‘m’ and introducing the quantity ‘p’ and , equation (B2) and (B4) can also be written as (B5) x + 2ζ p x + p 2 x = p 2 y + 2ζ p y ( t ) and (B6) u + 2ζ p u + p 2 u = − y
3

once either x or u has been determined the other may be computed from Eqn. ku the Eqn.(B2) and (B4) or between Eqn. For example. (B3) 4 . if the ground motion is specified as an acceleration history and we are interested in resulting spring force. Eqn. (B4) and (B6) would probably be the most convenient. if we are interested in absolute displacement and both y (t ) and y (t ) are specified.The choice between Eqn.(B5) and (B6) in a given problem depends on how the ground motion is specified and what response quantity x or u is interested in. (B2) and (B5) would be the most convenient to use. Clearly. On the other hand.

In terms of Duhamel’s integral. simply by replacing in the latter solution xst by y (t). c = 0. (B5) reduces to. m x + p 2 x = p 2 y (t ) (B7) This equation is same as the differential equation governing the motion of a fixed base system subjected to a force for which the associated static displacement x st ( t ) = y ( t ) (B8) The solution of equation (B7) can therefore be obtained from the solution for the force-excited system considered previously. Eqn. the solution may be written as t v0 sin p t + p ∫ y (τ ) s in ( p ( t − τ ) d τ x (t ) = xo c o s p t + (B9) p 0 5 .Absolute Displacement of Undamped System For undamped systems.

It is only necessary to replace the quantity (xst)o in the expression for the amplification factor xmax ( xst ) o by the peak value of the base displacement. In the other words. yo . force-excited systems presented previously can also be interpreted as spectra for the absolute displacement of baseexcited systems. 6 .The histories of the base motion and the excited force must naturally be the same in the two cases.It follows further that the response spectra for the fixed-base. the spectral ordinates should be interpreted to be ratio of xmax ( xst ) o .

Example: A vehicle. the vehicle is subjected to a base motion. Prior to crossing the irregularity. v yo L As it crosses the irregularity. idealized as a SDF undamped system moved at a speed of 20 m/s over an irregular rigid pavement. the vehicle is considered to have no vertical motion. the displacement history of which is a half sine pulse. The maximum displacement of the mass may then be from the response spectrum presented before. what would be the maximum vehicle displacement of the vehicle for (a) L=1. The shape of the irregularity is a half sine wave and its peak value is. 7 . If the natural frequency of the vehicle is f = 2 cps.5 m (b) L=6 m (c) L=24 m.

the values of frequency parameter.7 y0 X max = 1. for (a) for (b) for (c) X max = 1.1 m/s. for the three cases are for (a) for (b) for (c) ft1 = 0.6 y0 The corresponding values of X max are . is given by t1=L/V and that.1 y0 8 .15 ft1 = 0. ft1.Noting that the duration of the pulse. V=20. t1.6 ft1 = 2.4 X max = 0.

If the irregularity were a full sine wave and L the length of each half wave. the resulting displacements could be determined from the spectrum given before. for (c) L=24 m .2 y0 9 . The result in this case are as follows for (a) L=1.5 m.23 y0 X max = 1. X max = 0.58 y0 X max = 3. for (b) L= 6 m .

tan γ = 2ζ ω p = 2ζφ 10 .Absolute Displacement of Damped System The analogy referred to in the preceeding section is valid only for undamped systems. for the special case of a sinusoidal base excitation. y (t ) = y0 sin wt p2 ( y + the RHS of equation (B5) reduces to . It can be used as an approximation for damped systems only when the damping is small. However. 2 ⎛ω ⎞ 2ζ ω y) = p2 y0[sinωt + 2ζ cosωt] = p2 y0 1 + 4ζ 2 ⎜ ⎟ sin (ωt + γ ) p p ⎝ p⎠ where γ is a phase angle defined by .

the solution for steady state response may be written by analogy to the solution given by equation 69 for the corresponding force exited system. It is only necessary to interpret the quantity (xst)0 in the later solution ⎛ω ⎞ as.In this case. y0 1 + 4ζ 2 ⎜ ⎟ = y0 1 + 4ζ 2φ 2 ⎝ p⎠ 2 This leads to 1 + 4ζ 2φ 2 x (t ) = y0 (1 − φ 2 ) 2 + 4ζ 2φ 2 11 .

Spring deformation of systems subjected to Base excitation The deformation. (B4) and Eqn. x 12 ..my(t). fixed-base system.. u . of the base-excited systems can also be obtained from the equivalent force excited. Then the two equations will be similar. The initial conditions of u for the base excited systems are the same as those on force excited systems. (B1) . P(t)= . will be equal to the displacement X. the solutions of the differential equations will also be the same.of the force excited system shown: k c m . The desired deformation. force P(t) for the force the excited system can be taken as.my(t) . Comparison of Eqn. u.

we conclude that.defined as x(t ) ( IAF ) = ( xst )0 It is desirable to evaluate at this stage the counter part of this factor for the base-excited system. ( xst )0 y0 (B12) 13 .In the analysis of fixed base system. P0 −m y0 y0 ( xst )0 = = =− 2 k k p where y0.. is the absolute maximum acceleration of the base motion. x( t ) − p 2u( t ) ( IAF ) = = . Noting that .extensive use was made of the concept of instantaneous amplification factor.

For. B12.. − p 2u ( t ) y0 − p 2u ( t ) y0 . subjected to a rectangular pulse of amplitude ..Example : Evaluate the deformation of a SDOF undamped system . t ≥ t1 . = 1 − cos pt pt1 ⎡ t⎤ sin ⎢ p( t − 1 ⎥ 2 2⎦ ⎣ = 2 sin 14 .making use of Eqn. . y and zero.Thus initial values of u and u are zero The solution is obtained from the information presented before.. y and of x and x are Assume that the initial values of. y 0 and duration t1. . t ≤ t1 .

velocity and displacement histories of the base motion considered in this solution are shown. y (t ) t1 . the deformation of an undamped system can be expressed in terms of Duhamel’s integral as follows. u 1 u ( t ) = u o c o s p t + 0 sin p t − p p ∫ y (τ ) sin ( p ( t − τ ) d τ 0 t (B13) 15 . For an arbitrary base motion. This type of base excitation is of interest in the design of equipment in moving vehicles..The acceleration.but is clearly of no interest in the design of structures subjected to ground motions. y( t ) t1 y (t ) t1 .

Pseudo . A( t ) = p2u( t ) Thus .it can readily be shown that.acceleration p 2u ( t ) The quantity in equation B12 .equation B12 can also be written as. this identity is not valid for damped systems.the pseudo-acceleration. B1.will be referred to as the instantaneous pseudo acceleration of the system.. acceleration of the mass.and will be denoted by A(t)..A(t) is also equal to the absolute .which has units of acceleration. 16 . ( x st ) 0 y0 (B14) (B15) Referring now to Eqn.and A(t) should be looked upon merely as an alternate measure of the spring deformation.for undamped systems. However. x . ( IA F ) = x (t) − A(t) = .u(t).

thus A = p 2U (B16) The product of the mass m and the pseudo-acceleration.without regards to sign. Qmax.and will be denoted by A. Qmax = CW (B18) 17 .will be referred to as the pseudo-acceleration of the system. The absolute maximum value of A(t).without regards to sign will be referred to as the spectral value of u and will be denoted by U. It is sometimes convenient to express Qmax in the form .A represents the maximum spring force. indeed Q max k = kU = m U = mp 2U = mA m (B17) This may also be viewed as the equivalent lateral static force which produces the same effects as the maximum effects by the ground shaking.Spectral Quantities The absolute maximum value of u.

V is identical to . V = p U (B20) The maximum strain energy stored in the spring can be expressed in terms of V as follows: Emax = (1/2) (kU) U = (1/2) m(pU)2 = (1/2)mV2 (B21) Under certain conditions.and care should be excercised in replacing one for the other. 18 .B17 and B18 it follows that.Where W = mg is the weight of the system.However this is not true in general. From Eqn.or approximately equal to the maximum values of the relative velocity of the mass and the bays. which represents the number of times the system must be capable of supporting its weight in the direction of motion. The quantity C is the so called lateral force coefficient. C=A/g (B19) Another useful measure of the maximum deformation.U and the two quantities can be used interchangeably.defined as. U is the pseudo velocity of the system. that we need not go into here.

Tripartite Logarithmic Plot 19 .Presentation of results in alternate forms (a) In terms of U (b) In terms of V (c) In terms of A 3.Deformation spectra 1.Obtained from results already presented 2.

The effect of discontinuous acceleration inputs is considered later. It exhibits a hump on either side of the nearly horizontal central portion. A which may be . a n d y 0 It is assumed that the acceleration trace of the ground motion. materially greater than the values of y 0 . The high-frequency limit of the response spectrum for discontinuous acceleration inputs may be significantly higher than the value referred to above. 20 .General form of spectrum . are smooth continuous functions.and the information presented should not be applied to such inputs.and attains maximum values of .and hence the associated velocity and displacement traces. U. It approaches U=y0 at extreme left. y 0 .V and.. a value of A = y 0 extreme right.

The characteristics can be defined by reference to the response spectrum for the particular ground motion under consideration.Spectral regions The characteristics of the ground motion which control the deformation of SDF systems are different for different systems and excitations. 21 .

Spectra for maximum and minimum accelerations of the mass (undamped elastic systems subjected to a Half cycle Acceleration pulse) 22 .

Spectra for maximum and minimum acceleration of the mass (undamped Elastic systems subjected to a versed-sine velocity pulse)

23

Deformation spectra for undamped elastic systems subjected to a versedsine velocity pulse

24

‘B’ Level Earthquake (β=10% ; µ=1.0)

25

Deformation spectrum for undamped Elastic systems subjected to a half-sine acceleration pulse 26 .

All these values can be read off directly from the logarithmic plot. On such a plot . 27 . and diagonal lines extending downward from left to right represent constant values of A. with the abscissa representing the natural frequency of the system. Advantages: • The response spectrum can be approximated more readily and accurately in terms of all three quantities rather than in terms of a single quantity and an arithmetic plot. • In certain regions of the spectrum the spectral deformations can more conveniently be expressed indirectly in terms of V or A rather than directly in terms of U.V (in a dimensional or dimensionless form).f. From a single plot of this type it is thus possible to read the values of all three quantities. (or some dimensionless measure of it) and the ordinate representing the pseudo velocity .diagonal lines extending upward from left to right represent constant values of U.Logarithmic plot of Deformation Spectra It is convenient to display the spectra or a log-log paper.

D0 V Log scale y0 . Acceleration sensitive A0 Natural Frequency F (Log scale) General form of spectrum 28 ..Logarithmic plot of Deformation Spectra Velocity sensitive Displacement sensitive V0 y0 y0 .

Although it is of no interest in study of ground shock and earthquakes . X max = I mp t1 where I = ∫ P (t ) dt 0 29 . V = yo This result follows from the following expression presented earlier for fixed base systems subjected to an impulsive force. When plotted on a logarithmic paper.being the simplest form of acceleration diagram possible . the spectrum for the half sine acceleration pulse approaches asymptotically on the left the value.it is desirable to investigate its effect.Deformation Spectra for Half-cycle Acceleration pulse: This class of excitation is associated with a finite terminal velocity and with a displacement that increases linearly after the end of the pulse.

we conclude that. is U = yo or V = y p o ) 30 . yo . without regards to signs. X max = U and noting that o ∫ y(t) dt = y 0 t1 o yo or V = y U = p ( This result can also be determined by considering the effect of an instantaneous velocity change.i.Letting P(t ) = − m y (t ) and we obtain.e. uo = 0 and uo = − yo where. an acceleration pulse of finite magnitude but zero duration. u (t ) = − yo sin pt p The maximum value of u(t). The response of the system in this case is given by. uo u(t ) = uo cos pt + p sin pt Considering that the system is initially at rest.

y0 U= or V = y 0 p . where. u( t ) = − − u0 sin pt p The maximum value of u(t).we conclude that . 31 . instantaneous velocity change.the response of the system in this case is given by .without regards to signs. .is .(This result can also be determined by considering the effect of as .e an acceleration pulse of finite magnitude but zero duration. u0 = 0 and u 0 = − y 0 . . u0 u( t ) = u0 cos pt + sin pt p Considering that the system is initially at rest. y 0 i.

2 From the spectrum.Q0 ft1= 2 x 0. V y0 .5g. Evaluate also the equivalent lateral force coefficient C. Assume that y0 = 0.1sec. and the maximum spring force.f=2cps.Example: For a SDF undamped system with a natural frequency.1 = 0. 32 .U when subjected to the half sine acceleration pulse.t1=0.evaluate the maximum value of the deformation.

.1× 0.024 π 2 2 .4W ..81 = 0. π g 8t1 × 0..81 π π 2 1 0.one can start reading the value .5 .4 g Alternatively.5g = = 8× 0.5 = 0. 4π × × t 1× y0 2 . we find that A y0 33 ..1 1 U ≈ 2 × × × 9. 2 2 A 2π fV 2π × 2 × y0 = = C= = g g g Q0 = CW = 0.Therefore 1 2π fU ≈ f1 y0 = × 0.1× × 9. A = 0 . from the spectrum y0 proceeding this may.

.5 × 9.385 Q0 = 0.4W and U = A 0.Accordingly 1 A 0.025 34 .8 × 0. as read from the spectrum are y0 C = 0. determined is 0.7.4 C= = g g Q0 = 0.385W and U = 0.81 = = = 0. The value of V .. The exact value of . y0 y0 A approximate.024 m 2 2 2 2 4π × 2 p p . This leads to and A ..8 y0 0.8 × 2 × g = 0.

5 × 0.047 m 2 2 2 4π × 2 p 35 .5 g = = 0.5 × 9..75W C= U= A 1.81 = = 0.therefore. the results would be as follows ft1 = 2 × 0.1sec .5 × 0.75 g g Q0 = 0.75 = 1. .If the duration of the pulse were f1=0. = 1.75sec instead of 0.5 A 1.5 and A y0 .

as in the first case.1=1. the results would have been as follows: ft1=15 * 0.5 = 0.75 × 9.00082m U= 2 = 2 2 p 4π × (15 ) 36 .If the duration of the pulse were t1. A = 1.5 × 1.5 Therefore.75W and 0. but the natural frequency of the system were 15cps instead of 2cps.5 y A C = = 0.81 A = 0.75 g Q = 0.

f • The spectrum for the longer pulse will be shifted upward and to the left by a factor of 0.5 37 .• Plot spectra for inputs considered in the illustrative example and compare y0 For t1=0.10 = 7.75/0.1sec V y 0 Same as in both cases ..75sec y0 For t1=0.

5 (Log scale) 38 . this spectrum may be approximated as follows: (Log scale) =1.Design Spectrum May be determined from the spectrum by interpreting xmax A as y ( xst )0 When displayed on a logarithmic paper with the ordinate representing V and the abscissa f.

Deformation Spectra for Half-Cycle Velocity Pulses Refer to spectrum for β = 0 Note the following • • • At extreme right A = y0. In general for pulses of the same shape and duration with different n peak values 2 A = ∑ ( y0 ) j j =1 • If duration on materially different 39 .5 The peak value of A=2x1.6 y0 Explain why. Explain why? Frequency value behind which A = y0 is given by ftra= 1.

0 This relationship is exact when the maximum response is attained following application of the pulse. But it is valid approximately even when the peak responses occur in the forced vibration era. The maximum value of U is yo and the spectrum is bounded on the left by the diagonal line U =yo 40 . Improved estimate may be obtained by considering relative durations of the individual pulses and superposing the peak component effects.The peak value of V is about 1.6 yo It can be shown that the absolute maximum value of the amplification factor V y0 for a system subjected to a velocity trace of a given shape is approximately the same as the absolute maximum value of A0 y0 for an acceleration input of the same shape.be conservative.

It is insensitive to the shape of the pulse which can more clearly be seen in the acceleration trace. It follows that to smaller values of t1 / td corresponds to larger values of peak acceleration y0 41 . the limiting value of on the right is equal to These limits appear different in the figure because of the way in which the results have been normalised. In all cases. inclined portion of the spectrum to displacement sensitive. nearly horizontal region of the spectrum is governed by the peak value of the velocity trace. (b) The middle. (c) The right hand portion is clearly depended on the detailed features of the acceleration trace of the ground motion. Note that the abscissa is non-dimensionalised and the ordinate with respect to the total duration of the pulse and the ordinate with respect to the maximum ground velocity.It should be clear that. (a) The left-hand.

continuous input acceleration functions) 42 .Design Rules Design spectrum for the absolute maximum deformation of systems subjected to a half cycle velocity pulse (undamped elastic systems.

Deformation spectra for undamped elastic systems subjected to skewed versed-sine velocity pulses 43 .

U=y0. The system in this case is extremely flexible and the ground displacements is literally absorbed by the spring. β=0. (b) Peak value of V is approximately 3. (c) At the extreme left and of the spectrum. This would be expected. has two identical pulses. on the next page Note that: (a) The RHS of the spectrum is as would be expected from the remarks already made. 44 .2 yo.Deformation Spectra for Half-cycle Displacement Pulse See spectrum for undamped systems. as the velocity trace of the ground motion.

45 . Further more the peak value of U occurs at the same value of the dimensionless frequency parameter.However the spectrum is no longer bounded on the left by the line U= yo.f1 as the peak value of V. However it is necessary to interpret t1 as the duration of the displacement pulse. induced by a velocity input of the same shape. but exhibits a hump with peak value of U0=1.6 y0 It can be shown that the peak value of U/y0 for a system subjected to a displacement trace is approximately the same as the peak value of V/y0. rather than of that of velocity pulse.

Design Rules Design spectrum for maximum deformation of systems subjected to a half cycle displacement pulse 46 .

Deformation spectra for damped elastic systems subjected to a half cycle displacement pulse 47 .

.Deformation spectra for full cycle Displacement pulse The spectra on the next page are for the following full cycle displacement pulse : y yo t yo y . y 0.. 0..618yo 0. yo t .618yo t yo .94f1 yo f1 0... ..618yo . 48 .94f1 . 0. .

and is numerously greater than the peak value of the second pulse of the contribution of the first pulse.As would be expected .This corresponds to the first maximum.the maximum value of U in this case is approximately 3.2 yo .Furthermore. (b) The smooth transition curve which defines the second maximum. (c) The hump on the left.which occurs at approximately the instant that y attains its first maximum.This maximum occurs approximately at the instant that y(t) attains its second extremum. which corresponds to the maximum that occurs after termination of the pulse 49 . the left hand portion of the spectrum consists of three rather than two distinct parts: (a) The part on the extreme left for which U=yo .

Deformation spectra for elastic systems with viscous damping 50 .

and hence the associated velocity and displacement forces.Generation of results • General form of spectrum is as shown in next slide (a) It approaches V= y0 at the extreme left. it exhibits a hump on either side of the nearly horizontal central portion. y0 and y0 respectively. (b) It is assumed that the acceleration force of the ground motion. 51 . V and A. are smooth continuous functions. and attains maximum values of U. which may be materially greater than the values of y0 . value of A = y0 at the extreme right .

and the information presented should not be applied to such inputs.(c) The high frequency limit of the response spectrum for discontinuous acceleration inputs may be significantly higher than the value referred to above. The effect of discontinuous acceleration input is considered later . 52 .

General form of spectrum 53 .

Acceleration spectra for elastic system .El Centro Earthquake 54 .

SDF systems with 10% damping subjected to El centro record Base shear coefficient.secs 55 . C Building Code Natural period.

Systems the natural frequency of which corresponds to the Inclined left-hand portion of the spectrum are defined as low-frequency systems : systems with natural frequencies corresponding to the nearly horizontal control region will be referred to as a medium-frequency systems . The characteristics can be defined by reference to the response spectrum for the particular ground motion under consideration . 56 .Spectral Regions The characteristics of the ground motion which control the deformation of SOF systems are different for different systems and excitations. and systems with natural frequencies corresponding to the inclined right handed portion will be referred to as high-frequency systems.

Minor differences in these characteristics may have a significant effect on the magnitude of the deformation induced. Low frequency systems are displacement sensitive in the sense that their maximum deformation is controlled by the characteristics of the displacement trace of the ground motion and are insensitive to the characteristics of an associated velocity and displacement trace: Ground motions with significantly different acceleration and velocity traces out comparable displacement traces induce comparable maximum 57 .

for an excitation of a particular form. It follows that a system of a given natural frequency may be displacement sensitive.deformations in such systems. they are a function of the duration of the motion. • The boundaries of the various frequency regions are different for different excitations and. velocity sensitive or acceleration sensitive depending on the characteristics of the excitation to which it is subjected 58 .

59 . Explain high-frequency response to discontinuous functions.Effect of Discontinuous Acceleration Pulses The high frequency limit of the deformation spectrum is sensitive to whether the acceleration force of the ground motion is a continuous or discontinuous diagram. These spectra provide further confirmation to the statement made previously to the effect that low-frequency and medium-frequency systems are insensitive to the characteristics of the acceleration force of the ground motion. The limiting value given priority applies only to continuous A = y0 functions The sensitivity of the high-frequency region to the detailed characteristics of the acceleration input may be appreciated by reference to the spectra given in the following these pages.

60 .

Deformation spectra for damped elastic systems subjected to a full cycle displacement pulse 61 .

62 .

y and y The degree of periodicity for (the number of dominant pulses in) each diagram. y and y The predominant frequency (or deviation) of the dominant pulses in y .Application to Complex Ground Motions Compound Pulses Earthquake Records Eureka record El-centro record Design Spectrum Minimum number of parameters required to characterize the design ground motion Max values of y. Dependences of these characteristics on Local soil conditions Epicentral distance and Severity of ground shaking 63 .

. extreme high frequency ranges (A = y0).. .. 2u=A(t)=y(t) high frequency p A =y0 64 .... .Effect of damping: • Effect is different in different frequency ranges • Effect is negligible in the extremely low frequency regime (U=y0) and . . low frequency u=y(t) u0 =y0 . u+ p2u = y0(t) .

1954 S 11o E component.Eureka. California earthquake of Dec 21. 65 .

Eureka Quake 66 .

67 .

Elcentro .N-S component 68 .California Earthquake of May 18.1940.

cps 69 .V = pseudo velocity Yc Maximum Ground Velocity Undamped Natural Frequency. f.

pp 287-303) the following relationship is used. y0 is normally based on a statistical study of existing earthquake records. Nov 1973.32 m/sec : 1g for rock 0. y0. 70 .Further discussion of Design Response Spectra The specification of the design spectrum by the procedure that has been described involves the following basic steps: 1. 0.3 : 7. In the Newmark – Blume – Kapur paper (“Seismic Design spectra for Nuclear Power Plants”.9 : 14. αV. . between y0. ground velocity and ground displacement. Estimating the maximum spectral amplification factors. Jr. The relationship . αD. for the various parts of the spectrum.. αA . Estimating the maximum values of the ground acceleration. ASCE. of Power Division. Again these may be based on statistical studies of the respective spectra corresponding to existing earthquake records.6 m/sec : 1g for 2. The results will be a function not only of the damping forces of the system but also of the cumulative probability level considered.

86 1.99 1.08 1.66 3.37 1.92 2.12 1.20 1.73 2.46 2.01 1.31 2.38 2.01 1.Following are the values proposed in a recent unpublished paper by Newmark & Hall for horizontal motions: Damping %critical 0.1%) αD 3.59 2.65 1.30 2.04 2.36 1.71 2.85 1.10 4.26 αD Median (50%) αV 2.64 1.38 αV 3.69 1.01 .17 71 2.42 2.39 1.03 1.51 1.65 3.82 1.37 αA 5.21 2.38 3.89 1.24 2.08 αA 3.63 1.52 1.64 2.5 1 2 3 5 7 10 20 One sigma (84.29 1.24 2.74 2.84 3.84 1.

00127 8cps Note: In the spectra recommended in the Newmark-Blume-Kapur paper. Take the knee of amplified constant acceleration part of the spectrum at 8 cps and the point beyond which A = y0 at 25 cps /s 69 cm 30 x2.3 ye = 30 cm / sec and y0 = 25 cm. 72 .Example: Determine the response spectrum for a design earthquake with y = 0.3g C=0.3 = 30 yo =30cm/s A=0.3 Q=0. the line de slope upward to the left and the line of slopes further downward to the right.3 = 69cm/s 30 x2 .3N ζ Y=0. Take ζ = 0.05 use the amplification factors given in the preceding page.